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Let $f$ be an analytic function such that $f(z)$ is an element of $\mathbb R$ for all $z$ element of $\mathbb C$. Prove $f$ is constant.

Here's what I have done -

$f(z) = c + i0$, where $c$ is an element of $\mathbb R$

So i have component functions

$u(x,y) = c$, $v(x,y) = 0$

The partial derivative $u_x = 0$ and the partial derivative $v_x = 0$

The derivative, $f'(z)$ = $u_x +i v_x$, so I have $f'(z) = 0$

As $f'(z) = 0$ the function must be constant.

Does that seem right? One thing that I noticed when looking at question is that if $f(z)$ is an element of $\mathbb R$ then it is automatically constant...isn't that correct? But I would expect they are looking for more than that in an exam situation...

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I believe the question states $f(z) = f(x+iy) = u(x,y) + 0i$. The element in $\mathbb{R}$ could be different for different $z$. – Aryabhata Apr 29 '12 at 18:22
The question is not asking you to assume that the real part of the function is constant. It is asking you to assume only that the imaginary part is always $0$. – Antonio Vargas Apr 29 '12 at 18:22
Do you know any equations linking the real and imaginary parts of a differentiable function over $\mathbb C$? – Mark Bennet Apr 29 '12 at 18:28
@Aryabhata But doesn't $u(x,y) = c$ like I have done basically say the same thing you are saying. $c$ could be anything. – Jim_CS Apr 29 '12 at 18:32
@AntonioVargas I think I understand you mate. When it says f(z) is an element of $\mathbb R$, that element doesn't have to be a real number like, say 5, it could be, say, $2x +3y^3$. As $2x +3y^3$ has no imaginary part so will evaluate to a real number. Is that what you are getting at? – Jim_CS Apr 29 '12 at 18:35
up vote 5 down vote accepted

Solution 1: Let $g(z)=\dfrac{1}{f(z)-i}$. Then $g(z)$ is bounded and entire.

Solution 2: Let $h(z) =e^{i f(z)}$. Then $h(z)$ is entire and $\left| h(z) \right| =1$.

Both solutions rely on Liouville's theorem though....

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+1 I can't imagine a simpler proof than this, given Liouville's Theorem. – Alex Becker Apr 29 '12 at 19:10

Here is a pedestrian answer: Using the CR differential equations we have $$f'(z)=v_y(x,y)+i v_x(x,y)\equiv0\qquad(z=x+iy\in\Omega)\ .$$ Given any two points $z_0$, $z_1\in\Omega$ there is a differentiable curve $$\gamma:\ t\mapsto z(t)\in\Omega\qquad(0\leq t\leq1)$$ with $z(0)=z_0$, $z(1)=z_1$. Consider the auxiliary function $\phi(t):=f\bigl(z(t)\bigr)$. Then $\phi'(t)=f'\bigl(z(t)\bigr)\,\dot z(t)\equiv0$ and therefore $$f(z_1)-f(z_0)=\phi(1)-\phi(0)=\int_0^1\phi'(t)\ dt=0\ .$$

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When I was taught the theory of a complex variable, we started with the Cauchy-Riemann equations. Given that the complex part of $f$ vanishes, the partial derivatives of the real part of $f$ vanish. That seems to me to be the elementary approach indicated by the way the question is put.

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As the image of any non constant analytic map must be open and by open mapping theorem f is constant, I hope $f:\mathbb{C}\rightarrow \mathbb{C}$, a line i.e homeomorphic to $\mathbb{R}$ in $\mathbb{C}$ is closed set. Or you can argue by picards theorem that image of non constant entire function can omits atmost one complex number. and the most elementarily you start from cauchy riemann equation I am sure i will get both $u$ and $v$ constant and hence f is so by taking $f=u+iv$

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